THERMODYNAMICS FIRST LAW OF THERMODYNAMICS · PDF file03/05/2009 · FIRST LAW OF...

FIRST LAW OF THERMODYNAMICS CLOSED SYSTEM1

THERMODYNAMICS

FIRST LAW OF THERMODYNAMICS – CLOSED SYSTEM

It has been pointed out that energy can neither be created nor destroyed; it can only change forms. This principle is the first law of thermodynamics or the conservation of energy principle.

During an interaction between a system and its surroundings, the amount of energy gained by the system must be exactly equal to the amount of energylost by the surroundings.

Energy can cross the boundary of a closed system in the form of heat and work.

Heat Transfer

Heat transfer or just heat is defined as the form of energy that is transferred between two systems (or a system and its surroundings) by virtue of temperature difference. Hence, if two systems are the same temperature, there can be no heat transfer.

A process during which there is no heat transfer is called an adiabatic process. A process can be adiabatic if both the system and its surroundings are at the same temperature or if the system is well insulated.

An adiabatic process is not necessarily an isothermal process!

The amount of heat transferred from state 1 to state 2 is demoted Q12 or just Q, it has units kJ.

m

Qq (kJ/kg)

The heat transfer rate is given as

2

1

t

t

dtQQ (kJ)

When the rate of heat transfer remains constant during a process

tQQ (kJ)


Modes of Heat Transfer

There are three modes of heat transfer: conduction, convection and radiation.

Conduction heat transfer occurs in stationary solids, liquids and gasses. It occurs due to random molecular motion when atoms or molecules of the system collide. The heat transfer rate equation is called the Fourier’s lawgiven below.

Fourier’s law: dx

dTAkQ tcond (W)

x

TAkQ tcond

(W)

Where Δx is the constant thickness of the system, ΔT is the temperature difference across the layer and A is the area perpendicular to the heat flow. Finally, kt is called the thermal conductivity and it is a measure of a materials ability to transfer heat.

Convection heat transfer occurs between a solid surface and a moving fluid (liquid or gas). It occurs due to random molecular motion and bulk fluid motion. The first layer of fluid adjacent to the plate is stationary hence, heat transfer is due to conduction. Thereafter, it is due to hot fluid being carried away and replace by cooler fluid.

The heat transfer rate equation is called the Newton’s law of cooling given below.

Newton’s law of cooling: )( fsconv TThAQ (W)

Where h is the convection heat transfer coefficient, A is the surface area through which heat transfer takes place, Ts is the surface temperature and Tf

is the bulk fluid temperature.

Radiation is the energy emitted by matter in the form of electromagnetic waves. Unlike conduction and convection, it does not require the presence of an intervening medium. In fact it is fastest through vacuum.

Solids, liquids and gasses can emit, absorb, reflect or transmit radiation.


The maximum rate of radiation that can be emitted by a surface at Ts is governed by the Stefan-Boltzmann law as

4srad ATQ (W)

Where A is the area of the emitting surface and σ = 5.67 x 10-8 is the Stefan-Boltzmann constant. Only an idealized blackbody can emit this maximum amount of radiation at Ts. Real objects emit less and the rate of heat transfer is expressed:

4srad ATQ (W)

Where ε is the emissivity and measures the ability of an object to emit radiation. It ranges between 10 where ε = 1 is a blackbody

Absorbed radiation also affects the energy levels in a system it is given by

incabs QQ (W)

Where α is called absorptivity and it measure the ability of an object to absorb radiation.

In the special case of a small object of emissivity ε, a surface area A and absolute temperature Ts, is enclosed in a large surface at absolute temperature Tsurr. The net rate of radiation heat transfer is given as:

44surrsrad TTAQ (W)

Work

As mentioned, energy can cross a system boundary by heat transfer and work. Therefore, if the energy crossing the boundary of a closed is not heat, it is work. The amount of work done during a process between state 1 and state 2 is denoted by W12 or just W, it has units kJ.

The work done per unit mass is given as

m

Ww (kJ/kg)

The work done per unit time is called power.


Sign Convention for Heat and Work

Heat transfer into a system is positive.

Work done by the system is positive.

Heat transfer out of a system is negative.

Work done to the system is negative.

Electrical Work

When calculating rate of electrical work we use the expression:

VIWe (W)

Where We is the electrical power, V is the potential difference and I is the current.

In the case where V and I varies with time, the electrical work done during a time interval Δt is expressed:

2

1

VIdtWe (kJ)

When V and I are constant, expression above is reduced to

tVIWe (kJ)

Mechanical Forms of Work

We are familiar with the expression

FsW (kJ)

Where F is the force acting on a body and s is the distance traveled be the body.

If the force is not constant than

2

1

FdsW (kJ)

There are two requirements for work interaction between a system and its surroundings to exist:

1. There must be a force acting on the boundary and

2. The boundary must move.


Boundary Work

One form of mechanical work frequently encountered in practice is associated with the expansion or compression of a gas in a piston-cylinder device. During this process, part of the boundary moves back and forth. The expansion and compression work is called moving boundary work or just boundary work.

The boundary work is analysed for a quasi equilibrium process (or quasi static process), a process during which the system remains in equilibrium at all times. A quasi equilibrium process is achieved when the piston moves at low velocities. Under this condition, the system work output is maximum and the work input is minimum.

The differential boundary work done is

PdVPAdsFdsWb

The total boundary work done during the entire process is then

2

1

PdVWb

Positive result indicates boundary work output (expansion).

Negative result indicates boundary work input (compression).

The area under the process curve on a P-V diagram is equal, in magnitude to the work done during a quasi-equilibrium expansion or compression of a closed system.

bWPdVdAAArea 2

1

2

1


Polytropic Process

During expansion and compression process of real gases, pressure and volume are often related by PVn = C, where n and C are constants. This is kind of process is called polytropic process.

HencenCVP

Substituting into

2

1

PdVWb ,

2

1

11221

11

22

1 11 n

VPVP

n

VVCdVCVPdVW

nnn

b

Since nn VPVPC 2211

For an ideal gas where

mRTPV

The above equation can also be written

n

TTmRWb

112


Gravitational Work

Gravitational work is defined as work done by or against a gravitational force field. In a gravitational force field,

mgF

Where m is the mass of the body and g is the acceleration of gravity. Then the work required to raise this body from level z1 to z2 is

2

1

12

2

1

zzmgdzmgFdzWg (kJ)

I.e. the change in potential energy where z2 – z1 is the vertical distance traveled.

Acceleration Work

The work associated with the change in velocity of a system is called acceleration work.

maF

dt

dVa

dt

dVmF n

Hence

2

1

21

22

2

1 2

1VVmVdVmVdt

dt

dVmFdsWg

I.e. it is the change in kinetic energy of the body.


Shaft Work

A force F acting through a moment arm r generates a torque τ

rF

Fr

This force acts through a distance s where

nrs 2

The shaft work is then

nrn

rFsWsh 22

The shaft power is given by

nWsh 2

Where n is the number of revolutions.


Spring work

Work done by a spring is given as

FdxWspring

Where

kxF

And k is the spring constant given in kN/m.

Hence

21

222

1xxkWspring (kJ)

x1 and x2 are the initial and final displacement of the spring respectively

Continued…


The First Law of Thermodynamics

The first law of thermodynamics as stated previously, is the conservation of energy principle. In this section, we will use the first law of thermodynamics to relate heat, Q, work, W and, total energy, E.

Recall that total energy, E refers to the energy a system possesses.

PEKE UE

Energy Balance

With regards to the first law of thermodynamics, the net change (increase and decrease) in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during that process.

Hence

systemoutin EEE ENERGY BALANCE

Where

Ein = total energy entering the system

Eout = total energy leaving the system

Esystem = change in the total energy of the system

Energy Change of a System, systemE :

The energy change of a system is expressed as below

Energy change = Energy at final state – Energy at initial state

Or

12 EEEEE initialfinalsystem

Or simply

PEKEUE

(Recall that E = U + KE + PE)

Where


)(

)(2

1

)(

12

12

12

zzmgPE

VVmKE

uumU

Most systems encountered are stationary systems i.e. the do not change in velocity or elevation during a process. Hence the change in kinetic and potential energies are zero (KE = PE = 0)

Hence the energy balance is very often reduced to UE .

Mechanisms for Energy Transfer, Ein and Eout:

Energy can be transferred to a system in three forms: heat, work and mass flow.

1. Heat transfer, Q Heat transfer to a system (heat gain) increases the energy of the system and heat transfer from the system (heat loss) decreases the energy of the system.

2. Work, W Work transfer to a system (i.e. work done one a system) increase the energy of the system and work transfer from a system (i.e. work done by the system) decreases the energy of the system.

3. Mass flow, m When mass enters a system, the energy of the system increases because mass carries energy with it. Likewise, when mass leaves the system it takes out some of the energy of the system.

Hence, the energy balance can be written as

systemoutmassinmassoutinoutinoutin EEEWWQQEE ,,

The subscripts in and out denote quantities that enter and leave the system respectively. All six quantities on the right side of the equation represent “amounts”, and thus they are positive quantities. The direction of any energy transfer is described by the subscript “in” and “out”. Thus we do not need to adopt a formal sign convention.


On a per unit time basis the energy balance is

systemoutin EEE

On a per unit mass basis the energy balance is

systemoutin eee


For a closed system undergoing a cycle, the initial and final states are identical therefore

012 EEEsystem

- In terms of heat and work interactions

innetoutnet QW ,,

When solving a problem that involves an unknown heat or work interaction, it is common practice to assume

EWQ

Where we assume there is a heat input into the system and work is done by the system (work output). When a negative answer is obtained for Q and W, it simply means the assumed direction is wrong and should be reversed.

Questions

1. The radiator of a steam heating system has a volume of 20 L and is

filled with superheated vapour at 300 kPa and 250 °C. At this

moment both the inlet and exit valves to the radiator are closed.

Determine the amount of heat that will be transferred to the room

when the steam pressure drops to 100 kPa. Also show the process on

a P-v diagram with respect to saturation lines.

2. A 0.5 m3 rigid tank contains refrigerant-134a initially at 200 kPa and

40 percent quality. Heat is now transferred to the refrigerant until the

pressure reaches 800 kPa. Determine (a) the mass of the refrigerant in

the tank and (b) the amount of heat transferred. Also, show the

process on a P-v diagram with respect to saturation lines.


3. A well insulated rigid tank contains 5 kg of a saturated liquid-vapour

mixture of water at 100 kPa. Initially, three-quarters of the mass is in

the liquid phase. An electric resistor placed in the tank is connected to

a 240-V source and a current of 8 A flows through the resistor when

the switch is turned on. Determine how long it will take to vaporize

all the liquid in the tank. Also, show the process on a T-v diagram

with respect to saturation lines.

4. An insulated tank is divided into two parts by a partition. One part of

the tank contains 2.5 kg of compressed liquid water at 60 °C and 600

kPa while the other part is evacuated. The partition is now removed,

and the water expands to fill the entire tank. Determine the final

temperature of the water and the volume of the tank for a pressure of

10 kPa.

5. A piston-cylinder device contains 5 kg of refrigerant-134a at 800 kPa

and 60 °C. The refrigerant is now cooled at constant pressure until it

exists as a liquid at 20 °C. Determine the amount of heat loss and

show the process on a T-v diagram with respect to saturation lines.

6. A piston-cylinder device contains steam initially at 1 MPa, 350 °C,

and 1.5 m3. Steam is allowed to cool at constant pressure until it first

starts condensing. Show the process on a T-v diagram with respect to

saturation lines and determine (a) the mass of the steam, (b) the final

temperature, and (c) the amount of heat transfer.


7. A piston cylinder device initially contains steam at 200 kPa. 200 °C

and 0.5 m3. At this state, a linear spring is touching the piston but

exerts no force on it. Heat is now slowly transferred to the steam,

causing the pressure and the volume to rise to 500 kPa and 0.6m3,

respectively. Show the process on a P-v diagram with respect to

saturation lines and determine (a) the final temperature, (b) the work

done by the steam, and (c) the total heat transferred.

8. A piston-cylinder device initially contains 0.5 m3 of saturated water

vapour at 200 kPa. At this state, the piston is resting of a set of stops,

and the mass of the piston is such tat a pressure of 300 kPa is required

to move it. Heat is now slowly transferred to the steam until the

volume doubles. Show the process on a P-v diagram with respect to

saturation lines and determine (a) the final temperature, (b) the work

done during this process, and (c) the total heat transfer.

9. Two rigid tanks are connected by a valve. Tank A contains 0.2 m3 of

water at 400 kPa and 80 percent quality. Tank B contains 0.5 m3 of

water at 200 kPa and 250 °C. The valve is now opened, and the two

tanks eventually come to the same state. Determine the pressure and

the amount to heat transfer when the system reaches thermal

equilibrium with the surrounding at 25 °C.


Specific Heats

It takes different amounts of energy to raise the temperature of identical masses of different substances by one degree.

- We need about 4.5 kJ of energy to raise the temperature of 1 kg of iron to from 20 to 30 ºC

- It takes about 41.8 kJ of energy to raise the temperature of 1 kg liquid water from 20 to 30 ºC

It is desirable to have a property that will enable us to compare the energy storage capabilities of various substances. This property is the specific heat.

The specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by one degree.

In thermodynamics, we are interested in the kinds of specific heats:

- Specific heat at constant volume (Cv)

- Specific heat at constant pressure (Cp)

The Constant Volume Heating Of a Gas

Cv can be viewed as the energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant.

Let a mass of gas (m) be heated at constant volume such that it rises from T1 to T2 and its pressure rises from P1 to P2. The heat received my the gas is

12 TTmCQ v ------------- Eq. 1


Applying the first law energy balance, assuming KE = PE = 0and no work is done (Δν = 0)

Therefore:

systemoutin EEE

PEKEUWQ

UQ ------------- Eq. 2

Hence,

UTTmCv 12 ------------- Eq. 3

Rewriting Eq. 3 produces

uTTCv 12

Therefore

T

u

TT

uuCv

12

12

In differential form

tconsvv dT

duC

tan

P2

P1

2

1

P

VV1 = V2


The Constant Pressure Heating of a Gas

As before, for a constant pressure process, heat received by the gas is given by

12 TTmCQ p ------------- Eq. 4

The process on the P-v diagram is show below

Work done during the process is given by the area under the curve. Therefore

12 VVPW

1122 VPVPW ------------- Eq. 5

Applying the first law energy balance, assuming KE = PE = 0 and no work is done (Δν = 0)

Therefore:

systemoutin EEE

UWQ

1212

11122212

11221212

HHTTmC

VPUVPUTTmC

VPVPUUTTmC

WUQ

p

p

p

1212 hhTTC p ------------- Eq. 6

Rewriting Eq. 6 produces

P1 = P221

P

VV1 V2


T

h

TT

hhC p

12

12

In differential form

tconspp dT

dhC

tan

Specific Heat Relations for Ideal Gases

A special relationship between Cp and Cv for ideal gases can be obtained by differentiating the relation

RdTdudh

RTuh

Pvuh

We know that dTCdh p and dTCdu v , therefore

RdTdTCdTC vp

Therefore,

RCC vp

Another important ideal gas relation is the specific heat ratio, γ which is given as

v

p

C

C

γ = 1.4 for air at room temperature.


Internal Energy, Enthalpy, and Specific Heat Relations for Solids and Liquids

Specific Heat

A substance whose specific volume is constant is called incompressible substance – solids and liquids.

For incompressible substance (solids and liquids)

CCC vp

Internal Energy Changes

)( 12 TTCu av

Where Cav is the average C value at average temperature.

Enthalpy Changes

By differentiating Pvuh , we obtain

PdvvdPdudh

Since v is constant then dv = 0, yielding

vdPdudh

Integrating this equation yields

PvTCPvuh av

For solids, the term vP is insignificant so

TCuh av

For liquids, there are two possible cases

1) Constant pressure process (heater), (P = 0) therefore,

TCuh av

2) Constant temperature process (heater), (T = 0) therefore,

Pvh


The Polytropic Process of a Gas

A gas undergoing a polytropic process expand or contracts according to

CPV n

For a process between state 1 and state 2 is thennn VPVP 2211

We have also established that the work done during a polytropic process is given by:

n

VPVPW

1

1122

Since PV = mRT

n

TTmRW

1

12

The Combination of the Polytropic Law PVn = C and the Ideal Gas Equation of a Perfect Gas

The law CPV n will enable calculations to be made of the changes in pressure and volume which occur during a polytropic process. Combining this iwht the characteristic equation of a perfect gas will enable variations in temperature to be found.

Consider a polytropic process in which the state of a gas changes from P1, V1, T1 to P2, V2, T2.

By the polytropic law:nn VPVP 2211 ------------- Eq. 1


Hence:

n

V

V

P

P

1

2

2

1 ------------- Eq. 2

By the ideal gas equation,

2

22

1

11

T

VP

T

VP ------------- Eq. 3

22

11

2

1

VP

VP

T

T ------------- Eq. 4


Hence, substituting Eq 2 into Eq 4,

2

1

1

2

2

1

V

V

V

V

T

Tn

Or,

1

1

2

2

1

n

V

V

T

T------------- Eq. 5

Also, from Eq 4n

P

P

V

V1

1

2

2

1

------------- Eq. 6

Substituting Eq 6 into Eq 4,

n

n

P

P

T

T1

2

1

2

1

------------- Eq. 7

Combining Eq. 5 and Eq. 7

n

nn

P

P

V

V

T

T1

2

1

1

1

2

2

1


Questions:

1) 2 kg of gas, occupying 0.7 m3, had an original temperature of 15 ºC. If was then heated at constant volume until its temperature became 135 ºC. How much heat was transferred to the gas and what was its final pressure? Take Cv = 0.718 kJ/kg K and R = 0.287 kJ/ kg K.

2) A gas whose pressure, volume and temperature are 275 kPa, 0.09 m3 and 185 ºC, respectively, has its state changed at constant pressure until its temperature becomes 15 ºC. How much heat is transferred form the gas and the work done on the gas during the process? Take R = 0.287 kJ/ kg K. and Cp = 1.005 kJ/ kg K.

3) A gas whose original pressure and temperature were 300 kPa and 25 ºC, respectively, is compressed according to the law PV1.4 = C until its temperature becomes 180 ºC. Determine the pressure of the gas after it is compressed.

4) 0.675 kg of gas at 1.4 MPa and 280 ºC is expanded to four times the original volume according to the law PV1.3 = C. Determine

a) the original and final volume of the gas

b) the final pressure of the gas

c) the final temperature of the gas

Take R = 0.287 kJ/ kg K

5) 0.25 kg of air at pressure of 140 kPa occupies 0.15 m3 and form this condition it is compressed to 1.4 MPa according to the law PV1.25 = C. Determine

a) the change of internal energy of the air

b) the work done on or by the air

c) the heat received or rejected by the air

Take Cp = 1.005 kJ/ kg K and Cv = 0.718 kJ/kg K


The Adiabatic Process of a Gas

When dealing with the general case of a polytropic expansion and compression, the process followed a law of the form PVn = C.

Now the adiabatic process is a particular case of a polytropic process in which no heat is allowed to enter of leave the system

Consider an adiabatic process in which the state of a gas changes from P1, V1, T1 to P2, V2, T2.

Then,

12 TTmCU v

Also

11

121122 TTmRVPVPW

Where gamma is known as the adiabatic index.

From the polytropic law,nn VPVP 2211

n

nn

P

P

V

V

T

T1

2

1

1

1

2

2

1

And the characteristic equation

2

22

1

11

T

VP

T

VP

Applying the first law energy balance,

UWQ

But for an adiabatic process, Q = 0

UW

Hence,

1212

1TTmC

TTmRv

vCR

1

Hence,


v

v

v

v

C

RC

C

R

C

R

1

1

But R = Cp - Cv therefore,

v

p

C

C

γ for air is 1.4

Question:

A quantity of gas occupies a volume of 0.4 m3 at a pressure of 100 kPa and a temperature of 20 ºC. The gas is compressed isothermally to a pressure of 450 kPa and then expanded adiabatically to its initial volume. Determine, for this quantity of gas

a) the heat transfer during the compression

b) the change of internal energy during the expansion

c) the mass of the gas

Take n = 1.4 and Cp = 1 kJ/ kg K

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THERMODYNAMICS FIRST LAW OF THERMODYNAMICS · PDF file03/05/2009 · FIRST LAW OF...

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